Question

A force, $$\vec{F}$$ = (6 $$\hat{x}$$ + 3 $$\hat{y}$$ ) N, is applied to an object at a point whose position vector with respect to the pivot point is $$\vec{r}$$ = (5 $$\hat{x}$$ + 4 $$\hat{y}$$ + 3 $$\hat{z}$$ ) m. Calculate the torque (in units on Nm) created by the force about that pivot point.

A) T_x =

B) τ_y =

C) τ_z =

Answer 1

Solution The vector product of the torque is, Using vector determinant fornm x (0-9)-y(0-18)+z(15-24) The magnitude of the vector is, 1-9)(18) +(-9 V486 22.05 Nm The torque created by the force about the pivot point is, The magnitude the torque is 22.05 Nm